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Visualizing complex number powers

Learn how powers of complex numbers behave when you look at their graphical effect on the complex plane.

Connection between i2=1 and where i lives

We began our study of complex numbers by inventing a number i that satisfies i2=1, and later visualized it by placing it outside the number line, one unit above 0. With the visualizations offered in the last article, we can now see why that point in space is such a natural home for a number whose square is 1.
You see, multiplication by i gives a 90 rotation about the origin:
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You can think about this either because i has absolute value 1 and angle 90, or because this rotation is the only way to move the grid around (fixing 0) which places 1 on the spot where i started off.
So what happens if we multiply everything in the plane by i twice?
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It is the same as a 180 rotation about the origin, which is multiplication by 1. This of course makes sense, because multiplying by i twice is the same as multiplying by i2, which should be 1.
It is interesting to think about how if we had tried to place i somewhere else while still maintaining its characteristic quality that i2=1, we could not have had such a clean visualization for complex multiplication.

Powers of complex numbers

Let's play around some more with repeatedly multiplying by some complex number.

Example 1: (1+i3)3

Take the number z=1+i3, which has absolute value 12+(3)2=2, and angle 60. What happens if we multiply everything on the plane by z three times in a row?
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Everything is stretched by a factor of 2 three times, and so is ultimately stretched by a factor of 23=8. Likewise everything is rotated by 60 three times in a row, so is ultimately rotated by 180. Hence, at the end it's the same as multiplying by 8, so (1+i3)3=8.
We can also see this using algebra as follows:
=(2(cos(60)+isin(60)))3=23(cos(60+60+60)+isin(60+60+60) =8(cos(180)+isin(180))=8

Example 2: (1+i)8

Next, suppose we multiply everything on the plane by (1+i) eight successive times:
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Since the magnitude of 1+i is
|1+i|=12+12=2,
everything is stretched by a factor of 2 eight times, and hence is ultimately stretched by a factor of (2)8=24=16.
Since the angle of (1+i) is 45, everything is ultimately rotated by 845=360, so in total it's as if we didn't rotate at all. Therefore (1+i)8=16.
Alternatively, the way to see this with algebra is
=(1+i)8=(2(cos(45)+isin(45))8=(2)8(cos(45++458 times)+isin(45++458 times))=16(cos(360)+isin(360))=16

Example 3: z5=1

Now let's start asking the reverse question: Is there a number z such that after multiplying everything in the plane by z five successive times, things are back to where they started? In other words, can we solve the equation z5=1? One simple answer is z=1, but let's see if we can find any others.
First off, the magnitude of such a number would have to be 1, since if it were more than 1, the plane would keep stretching, and if it were less than 1, it would keep shrinking. Rotation is a different animal, though, since you can get back to where you started after repeating certain rotations. In particular, if you rotate 15 of the way around, like this
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then doing this 5 successive times will bring you back to where you started.
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The number which rotates the plane in this way is cos(72)+isin(72), since 3605=72.
There are also other solutions, such as rotating 25 of the way around:
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or 15 of the way around the other way:
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In fact, beautifully, the numbers which solve the equation form a perfect pentagon on the unit circle:
Solutions to z5=1

Example 4: z6=27

Looking at the equation z6=27, it is asking us to find a complex number z such that multiplying by this number 6 successive times will stretch by a factor of 27, and rotate by 180, since the negative indicates 180 rotation.
Something which will stretch by a factor of 27 after 6 applications must have magnitude A276=3, and one way to rotate which gives 180 after 6 applications is to rotate by 1806=30. Therefore one number that solves this equation z6=27 is
3(cos(30)+isin(30))=3(32+i12)=32+i32
However, there are also other answers! In fact, those answers form a perfect hexagon on the circle with radius 3:
Solutions to z6=27
Can you see why?

Solving zn=w in general

Let's generalize the last two examples. If you are given values w and n, and asked to solve for z, as in the last example where n=6 and w=27, you first find the polar representation of w:
w=r(cos(θ)+isin(θ))
This means the angle of z must be θn, and its magnitude must be Arn, since this way multiplying by z a total of n successive times will in effect rotate by θ and scale by r, just as w does, so
z=Arn(cos(θn)+isin(θn))
To find the other solutions, we keep in mind that the angle θ could have been thought of as θ+2π, or θ+4π, or θ+2kπ for any integer k, since these are all really the same angle. The reason this matters is because it can affect the value of θn if we replace θ with θ+2πk before dividing. Hence all the answers will be of the form
z=Arn(cos(θ+2kπn)+isin(θ+2kπn))
for some integer value of k. These values will be different as k ranges from 0 to n1, but once k=n we can note that the angle θ+2nπn=θn+2π is really the same as θn, since they differ by one full rotation. Therefore one sees all the answers just by considering values of k ranging from 0 to n1.

Want to join the conversation?

  • blobby green style avatar for user ejcabanban
    In example 4, z^6 = -27 , it says that it is suppose to rotate 180 degrees because of the negative sign? Can someone explain this further?
    (32 votes)
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  • blobby green style avatar for user ejcabanban
    In the powers of complex numbers question section their is a question asking to : Find the solution of the equation whose argument is strictly between 180 and 270
    1) z^5 = -243i

    Now in this math slide it says if their is a negative in the equation then the argument will be 180 + k(360) ... however in the question when I ask for a hint it says the argument is 270 + k(360) . Can some explain how they got 270?
    (15 votes)
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    • aqualine ultimate style avatar for user David
      So we have z^5 = -243.
      This is the same as z^5 = 243 (0 - 1*i).
      Let x be the argument.
      When is cos(x) = 0 and sin(x) = -1?
      (With help from the unit circle) this is when x = 270 + k * 360 degrees.
      Therefore, the argument is 270 + k * 360 degrees and we can write:
      z^5 = -243i = 243 (cos(270 + k * 360) + i * sin(270 + k * 360))
      and continue to solve the problem.
      (31 votes)
  • duskpin tree style avatar for user Radek
    In the example 2, can those points make a spiral?
    Can you describe Fibocnacci with complex powers?
    (14 votes)
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    • leaf green style avatar for user asd asd
      I have literally no knowledge about this, but a Google search shows these 2 pages from wiki:
      https://en.wikipedia.org/wiki/Golden_spiral
      https://en.wikipedia.org/wiki/Logarithmic_spiral

      So what I got from it is that in the Example 2, you could make a curve that connects the dots and that curve would be called a logarithmic spiral.

      You can also construct a "Golden" logarithmic spiral using the Golden ratio as your "growth constant" (which is just the ratio between the angle and the magnitude in the example).

      However, "Golden" spiral is only an approximation of a Fibonacci spiral (which is an illustration of Fibonacci series). This is because the ratio of adjacent terms in the Fibonacci series is NOT equal to the Golden Ratio, instead it approaches it as your terms approach infinity.
      (13 votes)
  • blobby green style avatar for user msamorin
    I'm having trouble understanding a specific piece of solving for the powers of complex numbers when it asks to solve for the argument in a specific range.

    Example: Find the solution of the following equation whose argument is strictly between 225​∘ and 315∘. Round your answer to the nearest thousandth.
    z^7=128i

    Under the hints, it says the argument is 90 + (360)K, K being any integer. How do I determine what to add to (360)K? Any further explanation is greatly appreciated!
    (10 votes)
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    • primosaur ultimate style avatar for user Jordan Cooper
      It sounds like you're asking how they got the 90 in that problem. If that's not your question, please let me know.

      The 90 is the angle (in degrees) of 128𝑖, because 128𝑖 is directly above the origin on the complex plane, or 90° clockwise from the positive real axis.

      If the equation to solve were 𝓏⁷ = -128, you would use the angle of ‒128, which is 180°. If the equation were 𝓏⁷ = 128 ‒ 128𝑖, you would use its angle, 315°.
      (4 votes)
  • blobby green style avatar for user Thomas Evans
    I don't believe that this was covered in the videos or articles, but if you are confused (as I was) about why the angles can just be added when exponentiating (cos(x) + isin(x)), try using FOIL on it by squaring it.

    Using double-angle identities yields the result: (cos(2x) + isin(2x)).
    (10 votes)
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  • piceratops seed style avatar for user Michelangelo
    i need an explanation to solve this: z^10 = i

    i understand every step, but i don't understand, solving for theta :(
    ex:
    10 ⋅ θ = 90 + k ⋅ 360​
    θ = 9 + k ⋅ 36
    Remember that θ is strictly between 120, degree and 180, degree.
    so 120 - 9 = 111 and 180 - 9 = 171. This multiply is 144 so θ = 153.

    someone can help me understand these steps?
    why it is 144??
    ​​
    thanks
    Byeee :)
    (3 votes)
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    • hopper jumping style avatar for user Nils Petter
      To say that θ is between 120 and 180 degrees, is the same as saying:

      120 < θ < 180, but you know that θ = 9 + k ⋅ 36, so we could say:

      120 < 9 + k ⋅ 36 < 180, we can solve this as we usually solve equations (to find k):

      120 - 9 < k ⋅ 36 < 180 - 9
      111 < k ⋅ 36 < 171
      111 / 36 < k < 171 / 36
      3.083 ... < k < 4.75, since k is an integer we must have k = 4

      Then we have: θ = 9 + k ⋅ 36 = 9 + 4 ⋅ 36 = 9 + 144 = 153.

      Hope it helps! :)
      (15 votes)
  • blobby green style avatar for user R C
    Can someone explain why if we are cubing Cos (60) we get 180 degrees? In the problem the radius of 2 is cubed (2^3). If you have cos (60)^3, you get 1800 degrees?
    (5 votes)
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  • blobby green style avatar for user Andrea Menozzi
    in this exercise the hint explains how to find the angle, but i don't understand why it subtract 90/7 from 225. I am lost.
    here is the exercise. and a piece of the Hint that confuses me.

    Find the solution of the following equation whose argument is strictly between 225 and 315.
    z^7 = 128i

    HINT
    Remember that θ is between 225 and 315 degree.
    Therefore, we need to find the multiple of 360/7 that is strictly within the range of
    225-90/7 = 1485/7 and 315-90/7=2115/7.
    This multiple is simply 1800/7, so θ = 270

    why 225-90/7 ? and how do you know that the multiple is 1800/7 ?

    anyone could explain this please?
    (6 votes)
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  • blobby green style avatar for user Shriram V
    Is it just me or is that the Fibonacci spiral for (1+i)^8?
    (4 votes)
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  • duskpin ultimate style avatar for user samueltucker22
    In example 2, it says:

    16(cos(360)+i sin(360))
    =16

    Even though cos(360) = 1,
    and i*sin(360) = 1i

    Therefore shouldn't 16(cos(360)+i sin(360))
    =16(1+1i)
    =16+16i?

    Sorry if I'm asking a dumb question :)
    (2 votes)
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